# [18. 4Sum](https://leetcode.com/problems/4sum/)
# 思路
和[3sum](./15.%203Sum.md)这题基本一样的，注意这题由于循环层数比较多所以如果能
在外层循环判断一下的话可以提前终止循环或跳过某次循环，见代码。   
时间复杂度O(n^3)

# C++
``` C++
class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>>res;
        int len = nums.size(), low, high, sum;
        if(len < 4) return res;
        sort(nums.begin(), nums.end());
        for(int i = 0; i < len - 3; i++){
            if(i != 0 && nums[i] == nums[i - 1]) continue;
            
            // 提前退出或跳过不可能的情况
            if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3] > target) break; // 最小的都比target大了，可以提前终止循环
            if(nums[i]+nums[len-3]+nums[len-2]+nums[len-1] < target) continue;
            
            for(int j = i + 1; j < len - 2; j++){
                if(j != i + 1 && nums[j] == nums[j - 1]) continue;
                low = j + 1;
                high = len - 1;
                while(low < high){
                    sum = nums[i] + nums[j] + nums[low] + nums[high];
                    if(sum < target) 
                        while(++low < high && nums[low] == nums[low - 1]) ; // 不断右移low指针
                    else if(sum > target) 
                        while(low < --high && nums[high] == nums[high + 1]) ; // 不断左移high指针
                    else{
                        vector<int>tmp = {nums[i], nums[j], nums[low++], nums[high--]};
                        res.push_back(tmp);
                        while(low < high && nums[low] == nums[low - 1]) low++;
                        while(low < high && nums[high] == nums[high + 1]) high--;
                    }
                }
            }
        }
        return res;
        
    }
};
```
